practical course of physical chemistry...koya university department of chemistry practical course of...

Koya University Department of Chemistry Practical Course of Physical Chemistry / 2nd Year

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Practical Course

Of

Physical chemistry

Prepared by

A.L. Karzan Abdulkareem Omar


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Contents:

Experiment (1): Determine the Density of Liquids and Solids.

Experiment (2) : Determine the surface tension of liquid by capillary rise method.

Experiment (3): Determine the viscosity of a given liquid by Ostwald's

viscometer.

Experiment (4): Determine the radius of a molecule by viscosity measurements.

Experiment (5): Determine the molecular weight of a high polymer by viscosity

measurements e.g. (polystyrene).

Experiment (6): Determine the specific and molar refraction of a given liquid by

Abbe refractometer.

Experiment (7): Determine molecular weight by boiling point elevation

Experiment (8): Determine Molecular Weight by the Dumas Method

Experiment (9): Determine the heat of solution of oxalic acid from solubility in

water at different temperature.

Experiment (10): Determine the solubility of benzoic acid in water at different

temperature and hence its heat of solution.


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Some basic concepts

In experiment of chemistry, mostly students require standard solution to study

physical properties.

Standard solution:

A solution of known concentration is called standard solution. It contains a known

weight of substance in a definite volume of solution. Generally concentrations are

expressed in terms of

i) Normality: It is the number of gram equivalents of solute present in one

liter of solution.

Number of gram equivalents

Normality (N) =

Liter of solution

ii) Molarity: It is the number of moles of solute present in one liter of

solution.

Number of moles

Molarity (M) =

Liter of solution

iii) Molality: It is the number of moles of solute dissolved in 1000 gms of

solvent.

Number of moles of solute

Molality (m) =

Wt. of solvent in kg (1000 gms)


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iv) Mole fraction: The mole fraction of a component in a solution is defined

as the number of moles of that component divided by the total number of

moles of all components in the solution. If a solute is dissolved in solvent

then

Moles of solute

The mole fraction of solute =

Moles of solute + Moles of solvent

Moles of solvent

The mole fraction of solvent =

Moles of solute + Moles of solvent

v) Volume fraction: If two liquids are miscible to form a solution then the

volume fraction can be determined as follows:

v1

Volume fraction of one (V1) =

v1 + v2

v2

Volume fraction of other (V2) =

v1 + v2


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vi) Weight fraction: It is the ratio of weight of solute to the total weight of

solution.

Wt. of solute

Wt. fraction of solute =

Total wt. of solution

Wt. of solvent

Wt. fraction of solvent =

Total wt. of solution

If weight fraction multiplied by 100 it is termed as percentage by weight.

vii) Percentage weight of solute by volume of solvent: It is the weight of

solute dissolved in 100 ml of solution: (Wt/V) and (Wt/Wt). also there is

percentage volume of solute by volume of solvent (V/V).

viii) Parts per thousand (P.P.T.): This is a method of expression

concentration of solute if solution is sufficiently dilute.

If 20 gm of can sugar is dissolved in 1000 gm of water (or 1000 ml) then

the resultant solution is said to be 20P.P.T. it can be expressed as

P.P.T = gm/1000 gm of solution.


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ix) Parts per thousand (P.P.m.): This is the best method of express

concentration of solute when solution is expected to be very dilute.

If 20 gm of can sugar is dissolved in one million gm of water (or 1000 ml)

then the resultant solution is called 20 P.P.m. it can be expressed as

P.P.m = gm/1000000 gm of solution.

x) Strength of solution: The strength of solution is always expressed in

terms of grams per liter.

If the concentration is given in terms of normality then

Strength = Normality x Equivalent wt.

And if the concentration is given in terms of normality then

Strength = Molarity x Molecular wt.

xi) Dilution: A solution is diluted when extra solvent is added. The

concentration of dilute solution can be calculated by using this equation

which known as Dilution law

M1 V1 = M2 V2 Or N1V1 = N2V2.


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Preparation of solution

1) Preparation solution from solid material

M x Vml x M.wt

Wt (gm) =

1000

For example:

prepare o.1 M of NaOH in 100 ml of distilled water.

M.wt of NaOH = 23+16+1= 40 gm/mol

V = 100 ml

M = 0.1 M

Wt = ?

M x Vml x M.wt 0.1 x 100 x 40

Wt (gm) = = = 0.4 gm

1000 1000

Take 0.4 gm of NaOH and dissolved in quantity of water, when it dissolved

completely add it into 100 ml standard flask and fill it up to the mark.


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For example:

Prepare o.1 N of Ca(OH)2 in 100 ml of distilled water.

Eq.wt of Ca(OH)2 = M.wt / 2 = 74/2 = 37 gm / mol

V = 100 ml

N = 0.1 N

Wt = ?

N x Vml x Eq.wt 0.1 x 100 x 37

Wt (gm) = = = 0.37 gm

1000 1000

Take 0.37 gm of Ca(OH)2 and dissolved in quantity of water, when it dissolved

completely add it into 100 ml standard flask and fill it up to the mark.


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2) Preparation solution from liquid material

For preparation of dilute solution from concentrate liquid can be use these

information. Density, % and M.wt or Eq.wt.

Density of liquid

Sp.gr =

Density of water at 4C

Density of water at 4C = 1, i.e. Sp.gr = Density

% is the percentage of concentrate liquid in bottle.

Sp.gr x % x 1000 Sp.gr x % x 1000

M = Or N =

M.wt Eq.wt

By using dilution law:

M1 V1 = M2 V2 Or N1 V1 = N2 V2


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For example:

Prepare 0.5 M and N of H2SO4 in 100 ml of distilled water. If you know density

of concentrate H2SO4 = 1.84 gm/ ml, M.wt= 98 gm/mol and % = 96.

Sp.gr x % x 1000 Sp.gr x % x 1000

M = for N =

M.wt Eq.wt

1.84 x 96/100 x 1000 1.84 x 96/100 x 1000

= for =

98 98/2

1.84 x 0.96 x 1000 1.84 x 0.96 x 1000

= for =

98 49

M = 18.02 for N = 36.04


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Now, for Prepare 0.5 M and N of H2SO4 in 100 ml of distilled water, using

dilution law.

M1 V1(conc) = M2 V2(dilute) for N1 V1(conc) = N2 V2 (dilute)

18.02 x V1(conc) = 0.5 x 100 for 36.04 x V1(conc) = 0.5 x 100

V1(conc) = 2.77 ml V1(conc) = 1.38 ml

Take calculated volume of concentrate liquid and put it in standard flask fill up to

the mark by distilled water.


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Experiment (1)

Determine the Density of Liquids and Solids

Apparatus:

Beaker, density bottle or Pycnometer, …etc.

Chemicals:

Irregular solid powder, Irregular and regular wood, benzene and water.

Theory:

Density is defined as the mass per unit volume of a substance, and it is a physical

property of matter. A physical property can be measured without changing the

chemical identity of the substance. Since pure substances have unique density

values, measuring the density of a substance can help identify that substance.

Density is determined by dividing the mass of a substance by its volume:

The units of density are commonly expressed as g/cm3 for solids, g/mL for liquids,

and g/L for gases.

Density is also an intensive property of matter. This means that the value of density

is independent of the quantity of matter present.


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Fig. 1 pycnometer or density bottle

A) Determine the density of water and benzene

Procedure:

1. By using the electronic balance, weight empty density bottle (W1) with it

caps.

2. Fill the density bottle up to neck by distilled water and weight it (W2).

3. Find the weight of water (W3).

4. Density bottle has known volume (V).

5. Calculate Density of water.

6. Repeat same steps to find the density of benzene.


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Calculation:

W3

Density =

V

W3(weight of liquid) = W2(weight of density bottle with liquid) – W1(weight of empty density bottle)

V is known volume of density bottle

B) Determine the density of an irregular solid

Procedure:

1. Weight the irregular solid on a balance (W).

2. Fill the Measuring Cylinder with Water to a known Volume (V1).

3. Add the irregular solid to cylinder, record the increase level of water (V2).

4. Find the volume of an irregular solid (V = V2 – V1).

5. Calculate the density of an irregular solid.

Fig.2 Irregular solid


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OR use a Eureka Can to find the Volume.

Procedure:

1. Weight of the irregular solid on a balance.

2. Add water until just overflowing.

3. Place a Measuring Cylinder under the spout.

4. Add the irregular solid to Eureka Can.

5. Collect the Water and read off the Volume.

6. Calculate density of an irregular solid.

Fig. 3 Eureka Can


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C) Determine the density of regular solid

Procedure:

1. Weight the regular solid on a balance.

2. Measure the three lengths and calculate the Volume.

(ie V = l x w x h )

3. Calculate the density of a regular solid.

Fig. 4 Regular solid


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Experiment (2)

Determine the surface tension of liquid by capillary rise method

Apparatus:

Capillary tube and beaker.

Chemicals:

Benzene, acetone and water.

Theory:

SURFACE TENSION [γ ] is the force per unit length that must be applied parallel

to the surface so as to counterbalance the net inward pull and has the units of

dyne/cm or N/m.

Figure. 5

Figure 5. (a) A molecule within the bulk liquid is surrounded on all sides by other

molecules, which attract it equally in all directions, leading to a zero net force. (b)

A molecule in the surface experiences a net attractive force pointing toward the

liquid interior, because there are no molecules of the liquid above the surface.


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Figure5. illustrates the molecular basis for surface tension by considering the

attractive forces that molecules in a liquid exert on one another. Part a shows a

molecule within the bulk liquid, so that it is surrounded on all sides by other

molecules. The surrounding molecules attract the central molecule equally in all

directions, leading to a zero net force. In contrast, part b shows a molecule in the

surface. Since there are no molecules of the liquid above the surface, this molecule

experiences a net attractive force pointing toward the liquid interior. This net

attractive force causes the liquid surface to contract toward the interior until

repulsive collisional forces from the other molecules halt the contraction at the

point when the surface area is a minimum. If the liquid is not acted upon by

external forces, a liquid sample forms a sphere, which has the minimum surface

area for a given volume. Nearly spherical drops of water are a familiar sight, for

example, when the external forces are negligible.

Figure.6 Figure.7

Cont. angle

water and glass


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When a capillary tube is placed in a liquid, it rises up the tube a certain distance.

By measuring this rise, it is possible to determine the surface tension of the liquid.

It is not possible, to obtain interfacial tensions using the capillary rise method.

i) Cohesive force is the force existing between like molecules in the surface of

a liquid

ii) Adhesive force is the force existing between unlike molecules, such as that

between a liquid and the wall of a glass capillary tube

When the force of Adhesion is greater than the cohesion, the liquid is said to wet

the capillary wall, spreading over it, and rising in the tube.

The upward force: Due to surface tension of the liquid at any point given by:

Upward force = γ 2 cos Ө r π

Where, γ is surface tension, r is radius and Ө = the contact angle between the

surface of the liquid and the capillary wall.

The downward force: Due to gravity force given by:

Downward force = g h d π r2 cos Ө

Where, g is acceleration, h is height of liquid in capillary tube and d is density of

liquid.

At Maximum height, the opposing forces are in equilibrium:

γ 2 cos Ө r π = g h d π r2 cos Ө

ɣ = ½ h g d r


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Procedure:

1. Clean a capillary tube with water and acetone, dry it.

2. Prevent bubbles to form in capillary tube.

3. Immersed part of a capillary tube in water bath.

4. Suck up water with help of rubber pipe which attached to capillary tube.

5. Water will be rise through capillary tube and stabilized at specific height.

Measure this high by ruler and record it.

6. Measure the surface tension of water at room temperature 25C, 30C and

35C.

7. Measure the surface tension of benzene at room temperature.

Calculation:

ɣ = ½ h g d r

ɣ is surface tension of liquid

h is capillary rise

g is acceleration

d is density of liquid

r is radius of capillary tube (0.045 cm).

Q/ Explain effect of temperature on surface tension of liquid.


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Experiment (3)

Determine the viscosity of a given liquid by Ostwald's viscometer.

Apparatus:

Ostwald's viscometer and stop watch.

Chemicals:

Benzene, acetone or any liquid.

Theory:

It is a general property of fluids (liquids and gases) to flow under an applied force.

When a liquid through a tube, a layer of the liquid in contact with the wall of the

tube remains stationary whereas the layer in the center has the highest velocity.

The velocity of different intermediate levels increases continuously with distance

from the wall of the tube to the center. Thus there is a movement of different layer

over one another in the direction of flow. This relative movement of different

layers experiences a frictional force and each layer experiences a frictional force

and each layer exerts a drag on the next layer in backward direction. This internal

friction or resistance which retards the flow of the liquid is known as viscosity.

Figure. A


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If two layer of a liquid are dr meters apart and have a velocity difference du m s-

1,which shown in Fiq.(a) then a force F acting per unit area of contact A m2 is

given by

The frictional force F, resisting the relative motion of any two adjacent layers, is

proportional to A, the area of the interface between them, and to du/dr, the velocity

gradient between them. This is Newton's Law of

Viscous Flow,

F A du / dr = A du / dr

is the coefficient of viscosity and is defined at the force per unit area required to

maintain unit different of velocity between two parallel layers in the liquid, unit

distance apart. It is expressed as dynes cm-2.

The common unit of are poise, centipoises (1/100 poise) and millipoise (1/1000).

The C.G.S unit of is (g.cm-1 sec-1).


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Factors affecting viscosity of liquids:

1) The viscosity of liquid depends upon the strength of intermolecular forces. If

the intermolecular force of attraction is more, the viscosity will be also

more.

2) The viscosity of liquids increase by increase molecular weight of a liquid.

3) Polar compound have more viscosity.

4) The viscosity of liquids increase by the presence of solute in it.

5) Liquid with large branch chain molecules have higher viscosity.

6) An increase in temperature decreases the viscosity of liquids.

7) The increase in pressure increases the viscosity to small extent.

Measurement of viscosity

The common method used for the measurement of viscosity is the observation of

flow of liquid through a capillary tube. Poisuille gave an expression for the

viscosity of liquid on the basis of its flow through capillary tube as

r4t

=

8Pvl

Where, is viscosity, r is the radius of the capillary tube, t is the time required for

the volume V of the liquid to flow through the length l, P is hydrostatic pressure on

the liquid.


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The simple method of comparison of the viscosities of two liquid is generally used.

The viscosity of a liquid is measured with respect to other standard liquid,

generally water. In this method the flow time of equal volumes of two liquids

through the same viscometer is measured. The apparatus generally used for the

measurement of viscosity is Ostwald's viscometer.

Ostwald's Viscosity

Poiseuill's apparatus modified by Ostwald is U-shape tube with two bulbs A and B

and a sort of fine capillary CD. The volume of liquid within EC is allowed to pass

through the fine capillary CD and time required to flow the liquid is noted down.

Bulb B works as reservoir to collect liquid. The force driving the liquid through the

capillary CD is equal to h x x g, where h is the mean difference of height

between the levels of the liquid in two limbs of the tube, is the density of the

liquid and g the gravitational constant. The force which oppose this flow depends

on dimensions of the capillary and viscosity of liquid. Thus the time of flow of

liquid from C to D in the capillary is directly proportional to the viscosity and

inversely proportional to the driving force. Thus

Fig.8 viscometer.


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1

For liquid 1, T1

h 1g

2

For liquid 2, T2

h 2g

T1 1 h 2g 1 2

= =

T2 2 h 1g 2 1

T2 2 1

2 =

T1 1

If the absolute viscosity of one liquid is known, that other can be calculated by

measuring T1,T2. 1 and 2.


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Procedure:

1) Wash and clean the viscometer with chromic acid and distilled water

thoroughly. Rinse it with acetone and dry with drier.

2) Take 10 ml of distilled water put it in the viscometer.

3) Suck up distilled water up to the mark and keep a stop watch ready.

4) Let off liquid, start the stop watch and note down the time required to flow

the water through the capillary tube.

5) Repeat it for 3 times and find out mean time (t1).

6) Take out water, rinse and dry it with acetone.

7) Take 10 ml of benzene put it in the viscometer.

8) Suck up benzene and note down time.

9) Repeat it for 3 times and find out mean time (t2).

10) Determine the density of benzene and water by using density bottle.

Calculation:

Calculate the viscosity of liquid (i.e. benzene) by using the formula.

T2 2 1

2 =

T1 1

Viscosity of water (1) is 8.91 x 10-3 poise.


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Experiment (4)

Determine the radius of a molecule by viscosity measurements.

Apparatus:

Ostwald's viscometer, stop watch, standard flask, etc.

Chemicals:

Sucrose solution (0.1M) and distilled water.

Theory:

The size of molecule can be determined by viscosity of its solution. According

to Einstein's equation:

= 1 + 2.5

o

Where, is the viscosity of solution, o is the viscosity of solvent and is the

volume fraction of solute. This equation is based on the assumptions that the

solute particles are rigid spherical bodies without any mutual attraction. The

size of molecule must be more than the molecule of solvent but it should be

sufficiently smaller so that it can pass through the capillary of viscometer. Then

equation modified as

- 1 = (r – 1)= 2.5 (4/3 r3) No C x 10-3 = 6.3 x 1021r3C

o


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Where, r = /o = t/to, t is flow time of solution and to is flow time of solvent,

r is radius of solute particle in cm, No is Avogadro's number, C is concentration

of solute in mole/litter. Thus when (r – 1) is plotted against C, it will be a

straight line passing through origin having a slope equal to 6.3 x 1021r3.

This experiment consists of the determination of viscosity of a series of sucrose

solution in concentration range of 0.005 to 0.1 M.

Procedure:

1) Clean the viscometer with acetone and dry it.

2) Prepare 0.02, 0.04, 0.06 and 0.08 M solution from stock solution (0.1M).

3) Determine the flow time for each solution and also for distilled water.

4) Determine the relative viscosity of each sucrose solution.

5) Plot a graph (r – 1) vs C.

Calculation:

From the straight line graph determine the slope and calculate the radius of the

sucrose molecule by using following equation:

Slope 1/3

r =

6.3 x 1021

Then convert r in cm to A.


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Experiment (5)

Determine the molecular weight of a high polymer by viscosity measurements e.g.

(polystyrene).

Apparatus:

Ostwald's viscometer, density bottle, pipette and stop watch.

Chemicals:

Polystyrene, acetone and distilled water.

Theory:

The average molecular weight of a polymer can be determined by simple viscosity

measurement and certain qualitative conclusions can be drawn about the general

form of macromolecules in solution. The viscosity of high polymer solution

depends on the size and shape of the molecule in solution. The measurement of

viscosity is a useful method in the study of polymer configuration.

The method involves the preparation of a series of the polymer solutions of

different concentrations in a suitable solvent and the measurement of their

viscosity by Ostwald's viscometer. If the absolute viscosity of the solution is and

that of the solvent is o the relative viscosity is given by r = / o

If the flow times for solution and solvent are t and to, then

r = t / to = / o

Provided the densities of solution and solvents are very close. Since the polymer

solutions used in the experiment are very dilute, their densities will be

approximately equal to that of solvent.


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t

Relative viscosity, r = =

to o

The specific viscosity sp = (r – 1)

Then the reduced viscosity will be given by

sp

red =

C

Where, C is concentration of polymer in gm/100 ml of solvent.

When the quantity sp/C for different solutions is plotted against concentration C,

a straight line will be obtained which on extrapolation to zero concentration given

the limiting viscosity [] which is related to the molecular weight of polymer by

Standinger equation as :

[] = KMa

Where, M is molecular weight of the polymer, K and a are the constants depend

upon the polymer and solvent at a particular temperature.


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The values of K and a for some polymer – solvent systems are given in table (1)

below at 25C.

Polymer Solvent K a

Polystyrene Toluene 3.7 x 10-4 0.62

Polyisobutylene Toluene 3.6 x 10-4 0.64

Polyvinyl alcohol Distilled water 2.0 x 10-4 0.76

Cellulose Acetone 1.49 x 10-4 0.82

Procedure:

1. Rinse viscometer with distilled water and then with acetone.

2. Prepare a stock solution (20 mg/ml) of polystyrene in toluene by weighing

out 5.000 g of polystyrene, transferring it to a 250 ml volumetric flask and

dissolving in toluene.

3. From stock solution, prepare16, 12, 8 and 4 mg/ml in 100 ml volumetric

flask and diluting with toluene up to the mark and keep them at same

thermostat.

4. Determine the time flow of solvent.

5. Determine the time flow of each solution.


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Calculation:

Calculate first the relative viscosity r = t/to and then specific viscosity

sp = (r – 1) for each solution. Then calculate the reduced viscosity sp/C, plot,

sp/C versus C. hence find out the intrinsic viscosity []. By referring the values of

K and from table for a given polymer, determine its average molecular weight.


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Experiment (6)

Determine the specific and molar refraction of a given liquid by Abbe

refractometer.

Apparatus:

Abbe Refractometer, dropper, cotton wool and density bottle.

Chemicals:

Alcohol or acetone and given liquid (Benzene)

Theory:

When a beam of light passes from rare to denser medium such as from air to a

glass or liquid, it bends towards the normal at the interface. This phenomenon is

known as refraction. According to Snell's law of refraction the ratio of the sine of

the angle of incidence and that of refraction is constant and is called as the

refractive index of liquid. The refractive index (n) of the liquid is given by

Sin i

n =

Sin r

Where, i is the angle of incidence and r is the angle of refraction.

The refraction index of a medium is also defined as the ratio of the velocity of light

in a vacuum to its velocity in given medium. Refractive index of the medium is

always greater than one.

Refractive index depends upon temperature and concentration. The refractive

index decreases with increase in temperature.


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However, the specific refraction defined by the following equation is independent

of temperature:

n2 – 1 1

r = x

n2 + 2

Where, r is the specific refraction, n is the refractive index and is the density of

liquid. The specific refraction is independent of temperature because the change in

the density of liquid.

Molar refraction or molecular refractivity of a substance is the product of specific

refraction and molecular weight of the substance. Thus, the molecular refraction

given by

n2 – 1 M

[R] =

n2 + 2

Where, M is the molecular weight of the substance and R is the molar refraction.

The molar refraction is additive and constitutive property. Each atom in a

compound has definite contribution to the molar refraction of compound. The

values of molar refraction are used in the determination of molecular structure.


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Procedure:

1) Place the refractometer on a table near a window so that sufficient light

reaches to prism.

2) Open the prism box by turning the lock nut. Clean the faces of both prisms

with help of cotton wool and alcohol or acetone and close the prism box

after drying.

3) Introduce few drops of liquid (Benzene) in the prism through the small hole

on the prism box, by means of dropper. A film of liquid will be enclosed

between the two prisms.

4) Focus the telescope by rotating the eye piece and adjust the mirror to reflect

maximum light into the prism box.

5) Move the prism box backward and forward until a clear boundary between

the light and dark regions appear.

6) Rotate the prism box until get the sharp boundary line.

7) Read the refractive index directly on the scale. Repeat the readings at least

three times and take the mean.

8) Determine density of the liquid by using density bottle.


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Calculation:

i) Determination of specific refraction (r).

n2 – 1 1

r = x

n2 + 2

ii) Determination of molar refraction (R).

n2 – 1 M

[R] = x

n2 + 2


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Experiment (7)

Determine molecular weight by boiling point elevation

Apparatus:

Beaker, capillary tube, test tube, thermometer, burner and rubber band.

Chemicals:

Paraffin oil and unknown liquid.

Theory:

The boiling point of a liquid is the temperature at which its vapour pressure

becomes equal to the atmospheric pressure.

This experiment demonstrates the use of colligative properties. The goal is to

measure the molecular weight of a non-volatile solute by determining the

concentration dependence of the boiling point elevation of a solution. The solvent

used must be one of the compounds commonly referred to as volatile; that is, it

must have an appreciable vapor pressure. One of the several useful aspects of

colligative properties is the fact that the vapor pressure of volatile solvents is

lowered when a non-volatile solute is used to make a solution. The difference

between the boiling points of solution and pure solvent at a certain constant

pressure is known as the elevation of boiling points of the solution. The result is

that such a solution will necessarily have a higher boiling point than that of the

pure solvent. The higher boiling point is due to the fact that a higher temperature is

needed in the presence of the non-volatile solute, which is not making any

contribution to the solution’s vapor pressure, in order to cause the volatile

component of the solution, the solvent, to exert one atmosphere of pressure. It must


38

be remembered that the boiling point elevation being investigated in this

experiment is a property of the solution as a whole and, for ideal dilute solutions, is

directly proportional to the solute concentration as shown in Equation

ΔTb = m • Kb

Where, m is the solution molality and Kb is the boiling point elevation constant

which is a function of the solvent not the solute. The value of ΔTb is the boiling

point of the solution minus that of the pure solvent, Tb* .

Table 2: boiling point elevation constants


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Factors affecting boiling point:

1. Pressure

By increase pressure increases boiling point.

2. Molecular weight

By increasing molecular weight increases boiling point.

3. Structure of the molecule & intermolecular interactions

a. Branching

Branching compounds have lower boiling point due to increases in

symmetry.

b. Polarity

Polar compounds have higher boiling point.

c. Van derWaals interactions

A compound which include Van derWaals interaction between their

molecules have lower boiling point.

d. H-bonding

A compounds which include H-bond have high boiling point due to H-

bond requires more energy to break down.

4. Impurities

A solution contain impurities have higher boiling point than the pure solvent.


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Procedure:

1. Place 2 ml of the unknown liquid in a small test tube.

2. Immerse a capillary tube sealed at the other end in the liquid.

3. Insert thermometer with capillary tube in to small test tube which contain the

unknown liquid or attach the outer tube to a thermometer by means of rubber

band.

4. Put test tube in an oil bath.

5. Start heating with a burner until a rapid stream bubbles coming out of the

capillary tube at this point record this temperature as the boiling point of the

liquid.

6. Repeat same procedure determine boiling point of solution.

Fig. 9 measuring boiling point


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Calculation:

Molecular weight can be calculated by using this equation:

ΔTb = m • Kb

ΔTb = T(solution) – T(pure solvent)

m = no. of moles/ 1 kg

Kb = molal constant


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Experiment (8)

Determine Molecular Weight by the Dumas Method

Apparatus:

Conical flask, heater, beaker (water bath), aluminum foil and electronic balance.

Chemicals:

Unknown volatile liquid and water.

Theory:

One of the early methods for the determination of the molar mass of volatile

substances was through the measurement of the density of the vapour of the

substance. The method is reliable and convenient and is still employed in some

situations .In this approach, the sample is added to a small flask, the flask is heated

and as the sample evaporates, the air is swept out of the container. Then flask is

cooled again, and the mass of liquid which condenses must be equal to the mass of

vapor that filled the flask in the previous step. A little skill is required to judge the

point at which the flask is just filled with sample vapour .

The combined gas law is given by the equation ;

p V = n R T, (1)

where p is the pressure of the gas in atmospheres, V is the volume of the gas

sample in litters, n is the number of moles of gas present, T is the temperature of

the sample in Kelvin, and R is the empirically determined quantity known as the

gas law constant which has the value of 0.0821 L atm mole1 deg-1This relationship


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describes very well the behaviour of gases at ordinary pressures and moderate

temperatures . In this experiment we rearrange the original combined gas law into

a form more convenient for our calculations. Recall that the number of moles of a

substance is equal to the mass of the substance divided by the mass of a mole of

that substance :

n = wt / M, (2)

Where wt represent the mass of the sample and M is the molar mass of the

substance .Substituting this relationship into the combined gas law :

pV = (wt/M) RT (3)

And rearranging to isolate the molar mass yields

M = (wt R T ) / (p V) (4)

You will recall that the ratio of mass to volume (wt/V) is commonly known as

density .

Thus equation 4 may be rearranged to :

M = (wt/ V) (R T / p) (5)

This is sometimes called the vapour density form of the combined gas law. The

basic outline of this experiment is to add a small amount of a liquid sample to a

small pre-weighed flask. The flask is then submerged into a boiling water bath.

As the sample evaporates, the air is swept out of the flask, and we finally have a

flask containing only the vapor of the unknown substance. If at that point the flask

is cooled, the vapors will condensed and their mass may be determined by

reweighing the flask. One may repeat the process and obtain an average of

replicate measurements. Subsequent measurements are the made to determine the


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volume of the flask, the temperature of the boiling water bath, and the prevailing

atmospheric pressure. Those values and equation 5 are used to calculate the molar

mass of the unknown liquid .

Procedure:

1) Determine the mass of the DRY flask to the nearest milligram.

2) Add about 5 mL of your unknown to the flask. Cut a small square of

aluminum foil and wrap over the top to the flask. Make a small hole in the

foil with a needle.

3) Immerse the flask in the boiling distilled water bath using a clamp and ring

stand. Be sue the flask is submerged at least to the neck.

4) Heat the flask until you no longer see a Schlieren pattern emerging form the

hole in the foil. Schlieren patterns are like the watery lines seen rising from

a heated surface .Remove the flask when the pattern disappears. The flask

should contain no liquid at this point. (Some people find that holding a

paper towel or shiny surface over the hole can be helpful. When vapor no

longer condenses on the towel or shiny surface, vapor is no longer emerging

and no liquid should remain in the flask ).

5) Cool the flask to room temperature and the vapors will condense into a small

amount of liquid .

6) Dry the exterior of the flask with a paper towel and weigh the flask and

contents to the nearest milligram. Remove the foil just immediately before

weighing .

7) Repeat the vaporization and condensation steps (2-6) twice more so that you

have three values for the mass of the condensed vapor .


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8) Obtain the atmospheric pressure from the barometer in the laboratory .

9) Determine the volume of the flask by filling it completely with water and

then measuring the volume of water contained in the flask with a large

graduated cylinder.

The temperature of the boiling water bath can be interpolated from the following

Table (3) :

11) From the average mass of condensed liquid and the temperature, volume and

pressure data, calculate the molar mass of your unknown liquid.

Fig. 10 Apparatus of Dumas method


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Calculation:

The equation 5 is used to calculate the molar mass of the unknown liquid .

M = (wt/ V) (R T / p)

M is molar mass (molecular weight).

Wt is weight of unknown liquid

V is volume of conical flask

R is 0.082 L atm mol-1

T is boiling point temperature of water bath in Kelvin

P is pressure

D is density of unknown liquid (0.7894 gm/ml)


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Experiment (9)

Determine the heat of solution of oxalic acid from solubility in water at different

temperature.

Apparatus:

Four beakers, pipette, cotton or filter paper, water bath, electronic balance and

oven.

Chemicals:

Oxalic acid and distilled water.

Theory:

Solubility is the property of a solid, liquid, or gaseous chemical substance called

solid to dissolve in liquid solvent to form a homogeneous solution of the solute in

the solvent. The solubility of a substance fundamentally depends on the used

solvent as well as on temperature and pressure. The extent of the solubility of a

substance in a specific solvent is measured as the saturation concentration where

adding more solute does not increase the concentration of the solution. Thus, in

acquiring the solubility of oxalic acid, the equivalent weight of oxalic acid from

the saturated sample of the solution is determined.

The Vant Hoff's equation also known as the Vukancic-Vukovis equation in

chemical thermodynamics relate the change in temperature (T) to the given the

enthalpy change (H) for the process. It follows that


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ln (S/T) = H / RT2

Final form of above equation will be

log S = - H / 2.303RT

Where, S is the solubility (mol/kg), T is the temperature in (K) and H is the heat

of solution (J/mol).

Procedure:

1. Take four clean and dry beakers and label from 1 to 4.

2. Weight empty beakers and record their masses.

3. Prepare 50 ml of a saturated solution of oxalic acid at 50C.

4. Pipette out 5 ml of saturated solution in beaker. To prevent sucking of small

crystals into pipette along with solution a small piece of filter paper or cotton

wrapped around the tip of pipette.

5. Weight beaker with solution and record it mass.

6. Put solution in oven to dry for (5-10) minutes at 100C and weight it after

cooled.

7. Repeat same procedure at 40C, 30C and 20C.


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Observation tabulate:

No.Temperature

(TC)

Massof

emptybeaker

(g)

Mass ofbeaker

andsolution

(g)

Mass ofsolution

(g)

Massof

beakerand

solid(g)

Massof

solid(g)

%Solubility

(S)

logS

1 50

2 40

3 30

4 20

Calculation:

Mass of solid

1. Solubility = x 100

Mass of solution

2. Plot log S against 1/T, H = - slope x 2.303R


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Experiment (10)

Determine the solubility of benzoic acid in water at different temperature and

hence its heat of solution.

Apparatus:

Beakers, burette, pipette, filter paper, funnel and thermometer ..etc.

Chemicals:

Solid benzoic acid, 0.05N NaOH solution and phenolphthalein indicator.

Theory:

The process of dissolution of solid into liquid is usually accompanied with the

absorption or evolution of heat. The heat of solution in the present experiment is

the heat evolved or absorbed when one mole of the solid is dissolved in a solution

which is already saturated. It differs from the heat of a solution at infinite dilution

by an amount equivalent to the heat of dilution from saturation to infinite dilution.

The effect of temperature on solubility is given by vant Hoff's equation:

d ln S H

=

dT RT2

Where, S is solubility (strength) and H, heat of solution.

On integration of the above equation, we have


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- H

Log S =

2.303 RT

Where, R = 1.987 cal/mole.

Procedure:

1) Prepare the saturated solution of benzoic acid at 45C, 35C, 30C, 25C,

20C and 15C.

2) By taking 100 ml of distilled water in a beaker. Add increasing amount of

benzoic acid with constant stirring until a small amount of solid remains

undissolved.

3) Pipette out 10 ml of saturate solution in a conical flask. To prevent sucking

of small crystal into the pipette along with the solution, a small piece of filter

paper wrapped around the tip of pipette and fastened with thread. The filter

paper should be removed before draining pipette.

4) Titrate this solution against 0.05N NaOH solution using ph.ph as an

indicator.

5) Then determine the normality and strength of benzoic acid.


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Calculation:

i) Strength = Normality x Equivalent weight

ii) Plot log S vs 1/T. from the slope of the straight line calculate heat of

solution H as

H = - 2.303 x R x Slope (cal/mole)


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